Question

A series L-C-R circuit is made by taking R = 100 $\Omega$, L=$\frac{2}{\pi }H$, C = $\frac{100}{\pi }\mu F$ This series combination is connected across an AC source of 220 V, 50 Hz. Calculate
(i) the impedance of the circuit and
(ii) the peak value of the current flowing in the circuit
(iii) the power factor of this circuit and compare this value with the one at its resonant frequency.

Answer 1

Given : $R=100\Omega$ $L=\frac{2}{\pi }H$ $C=\frac{100}{\pi }\mu F$ $V_{rms}=220V$ $f=50Hz$

(i) : $X_L=2\pi fL=2\pi \times 50\times \frac{2}{\pi }=200\Omega$

$X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 50\times \left(100/\pi \right)\times {10}^{-6}}=100\Omega$

Impedance of circuit $Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{{100}^2+(200-100{)}^2}=100\sqrt{2}\Omega$

(ii) : Peak value of voltage $V_o=\sqrt{2}{V}_{rms}=220\sqrt{2}V$

Peak value of current $I_o=\frac{V_o}{Z}=\frac{220\sqrt{2}}{100\sqrt{2}}=2.2A$

(iii) : Power factor $\cos \varphi =\frac{R}{Z}=\frac{100}{100\sqrt{2}}=\frac{1}{\sqrt{2}}$

Resonant frequency $f_r=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\frac{2}{\pi }.\frac{100}{\pi }}}=\frac{1}{20\sqrt{2}}$

So, $\frac{\cos \varphi }{f_r}=\frac{1/\sqrt{2}}{1/20\sqrt{2}}=20$